IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => IGCSE/ GCSE => Math => Topic started by: Padapop on June 04, 2009, 06:05:41 am
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Coming up tmmr.
It's almost the 24 hours since i've done it. Thought paper 1 was relatively hard, i got all of the questions except for question 9 and 12. 12 was hell for me :P
What do u think would come up for the next paper? (ps. Please tell me when the 24 hour mark is :D)
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erm guys! :S :S my teacher told us that approximate change and increments are not coming in the exam, i looked at the syllabus properly, and they are there... i have no clue how to solve approximate change and small/large increments
Can anyone help?
i know that it is an application of differentials...
and i looked up on the net
the
the change in y = dy/dx * change in x
is that right?
anything i need to know?
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erm guys! :S :S my teacher told us that approximate change and increments are not coming in the exam, i looked at the syllabus properly, and they are there... i have no clue how to solve approximate change and small/large increments
Can anyone help?
i know that it is an application of differentials...
and i looked up on the net
the
the change in y = dy/dx * change in x
is that right?
anything i need to know?
Yep thats right/ Really nothing else u need to know. Change in y = dy/dx * change in x
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hey, i am not sure but i think dat is not it!!
a change/increase in y=dy/dt
as it is over time.
then they ask u to find out the change in y
but they give u the change in x!!!
so next u equate
as change in x=dx/dt
dy/dt=dy/dx x dx/dt
we differentiate 2 find dy/dx and then put it in the equation
^^^^^^^^
and we get the change in y over time!!!!
tada!!
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smashbros, what your talking about it CONNECTED RATE OF CHANGE
we were finding the approximate change
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i think the person above was correct. its
change in x/change in y = dx/dy and you rearrange
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rate of change
the rate of change of a variable x with respect to time t is dx/dt
if x and y are related by the equaton y=f(x). then the rate of change dx/dt and dy/dt are related by
dy/dt = dy/dx * dx/dt
if its small change
then
&y/&x = dy/dx
the percentage change in x is &x/x *100
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hope this helped
:D
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for e.g (m = pi)
the area of a circle increases at a rate of 2m cm^2/s. Calculate the rate of increase of the radius at radius = 6cm
Ans: dA/dr = dt/dr * dA/dt
A= mr^2 therefore dA/dr = 2mr
at r = 6cm 2mr = 12m
so dA/dr = 12m
dA/dt = 2m cm^2/s
so the equation is
12m = dt/dr * 2m
therefore dt/dr = 12m/2m = 6
dt/dr = 6
therefore dr/dt = 1/6 (inverse)
easy..
hope it helped
thank u
note: m = pi (i cud not put the symbol of pi in the question)
:)
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Thanks :D + rep